Let the following nested radical :
$$\sqrt{665+2x}=x$$
There is a hidden quadratic equation and the result is :
$$x=\sqrt{666}+1$$
So we see the number $666$ appear .
Do you know (more or less trivial) nested radical where the devil's number appear ?
Can we build it as in my example ?
Thanks in advance and have fun .
Ps: I don't speak about Galois's theory .
On
Setting aside the fact that "the Number of the Beast" might be 616, or perhaps both 616 and 666 based on different spellings of "(Emperor) Nero", we can render 666 from an appropriate combination of the seemingly innocuous numbers 3 and 15:
$\sqrt{(15+\sqrt3)\sqrt3}$
To wit:
$\sqrt{(15+\sqrt3)\sqrt3}=\sqrt{(15\sqrt3)+\sqrt9}$
Then apply the identity
$\sqrt{a+\sqrt b}=\sqrt{(a+\sqrt{a^2-b})/2}+\sqrt{(a-\sqrt{a^2-b})/2}$
where here $a^2=675$ and $b=9$:
$\sqrt{(15+\sqrt3)\sqrt3}=\sqrt{(15\sqrt3+\sqrt{666})/2}+\sqrt{(15\sqrt3-\sqrt{666})/2}$.
The number 616 actually has this beat. Start with just $2\sqrt7$ and use the same identity as above, with $25^2=625$:
$2\sqrt7=\sqrt{25+\sqrt9}=\sqrt{(25+\sqrt{616})/2}+\sqrt{(25-\sqrt{616})/2}$.
I wrote a small program looking for equations of the form $ax^2 + 2bx + c = 0$ and found these quadratic equations below. Their roots all contain the number $\sqrt {666}$.
My program requires that $|b| \le 20$ and $|a| \le 700,\ |c| \le 700$.
Of course these limits can be extended.
By the way, your equation happens be to number 66 in the list below.
Equations 64 and 68 also look interesting to me since their solutions don't have a denominator.
In case anyone wants to play further with it, here is the Python code.